JEE Physics 2026 Question with Solution

Given: Masses are $m_1 = 15 \, \text{kg}$ and $m_2 = 25 \, \text{kg}$. Initial velocities are $u_1 = 10 \, \text{m/s}$ and $u_2 = -30 \, \text{m/s}$ because the spheres move in opposite directions. Specific heat is $s = 31 \, \text{cal/kg}\cdot{}^\circ\text{C}$ and $1 \, \text{cal} = 4.2 \, \text{J}$.

Find: The rise in temperature $\Delta T$.

In a perfectly inelastic collision, kinetic energy is lost. This lost energy is converted into heat. We first use conservation of momentum to find the common final velocity, then equate the kinetic energy loss to heat gained.

Using momentum conservation:

$$15(10) + 25(-30) = (15+25)v$$ $$150 - 750 = 40v$$ $$v = -15 \, \text{m/s}$$

Initial kinetic energy:

$$\frac{1}{2}(15)(10)^2 + \frac{1}{2}(25)(30)^2 = 750 + 11250 = 12000 \, \text{J}$$

Final kinetic energy:

$$\frac{1}{2}(40)(15)^2 = 20 \times 225 = 4500 \, \text{J}$$

Heat produced:

$$Q = 12000 - 4500 = 7500 \, \text{J}$$

Convert specific heat into SI units:

$$s = 31 \times 4.2 = 130.2 \, \text{J/kg}^\circ\text{C}$$

Now use calorimetry:

$$Q = (m_1+m_2)s\Delta T$$ $$7500 = (40)(130.2)\Delta T$$ $$\Delta T = \frac{7500}{5208} \approx 1.44^\circ \text{C}$$

Therefore, the rise in temperature is $1.44^\circ \text{C}$. The correct option is C.